Obections to Cantor's Theory (Wikipedia article)
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From: Robert Low on 30 Jul 2005 04:31 Dave Rusin wrote: > In article <b779a$42e9e8d1$82a1e3ad$17135(a)news2.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > > >>Under the new rules {{}} = {} . > > > Oh, that's just great. Tell me, what is the cardinality of the power > set of, say {1,2} ? I would say there are four elements: > P(S) = { {1,2}, {1}, {2}, {} } From his 'axiom', it looks as if the power set of any set is the same set again, since you effectively throw away the brackets: P(S) = {{1,2},{1},{2},{}}={1,2,1,2}={1,2} Of course, with the usual von Neumann definition of 1 and 2 this all collapses down to {}, which is presumably nothing at all, since ={}.
From: Martin Shobe on 30 Jul 2005 09:14 On Fri, 29 Jul 2005 14:15:27 +0200, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: >Martin Shobe wrote: >> On Fri, 29 Jul 2005 11:14:56 +0200, Han de Bruijn >> <Han.deBruijn(a)DTO.TUDelft.NL> wrote: >> >> >>>Martin Shobe wrote: >>> >>> >>>>On Thu, 28 Jul 2005 15:56:32 +0200, Han de Bruijn >>>><Han.deBruijn(a)DTO.TUDelft.NL> wrote: >>>> >>>> >>>>>But without the claim that it >>>>>is the one and only foundation possible. Why not have _several_ pillars >>>>>that provide a foundation, instead of just one? >>>> >>>>Actually, I don't have a problem with that. In a sense, we have that >>>>now with set theory and category theory (And lets not forget logic). >>>>But while physics will continue to provide inspirition to mathematics, >>>>it will not qualify as a foundation for methematics. >>> >>>And I don't want that either. Read my lips: >>> >>> A little bit of Physics would be NO Idleness in Mathematics >>> >>>See? Just that tiny pinch of salt in your otherwise tasteless soup. >>> >>>But nevertheless: *IN* your soup. >> >> >> A "tiny pinch of salt"? From the axioms I've seen you proposing that >> would be more like a ten pound bag of rock salt added to single >> serving. > >No, no. Nobody would like the soup anymore. Just a pinch. Nobody likes the soup anymore. The axiom that you seem most likely to want to add (Ax x = {x}), has as a consequence that Ax Ay (x e y) => (x = y). (It follows solely from the axiom and the definition of {x}). After that, you might as well not have sets at all. Martin
From: Torkel Franzen on 30 Jul 2005 09:37 Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > So if you replace ZFC by the PA system then you can still formulate the > theorem, but not "prove" it. Is that right? Goodstein's theorem is formalizable in the language of arithmetic and is provable in ZFC but not in PA, right. What is it you find puzzling about this?
From: Jesse F. Hughes on 30 Jul 2005 17:18 Vaughan Pratt <pratt(a)cs.stanford.edu> writes: > While I'm really enjoying this great thread, at the same time I'm > really bothered that no one has mentioned what to me is by far the > biggest problem with Cantor's theory, namely its claim to > universality. This is certainly the most thoughtful post in this thread. It is a symptom of Usenet, I guess, that it has generated only one other response that I see. The first part of the post (regarding V_4, etc.) seems to me to be the usual structuralist argument that set theory somehow misses the essence by forcing one to pick representations. Something like Benacerraf's point in "What numbers cannot be", right? The second part is new to me. It seems to be something like this: 2^A is naturally structured. It is a Boolean algebra. Thus, 2^- should be viewed as a functor from Set to BoolAlg. But then 2^(2^-) only makes sense if the outer 2^- is a functor with domain BoolAlg, and hence should take Boolean algebras B to the set of *Boolean* homomorphisms of B to 2. Here, I guess, you naturally regard 2 as the Boolean algebra (0 < 1). You then deduce that for finite sets A, 2^(2^A) "should" equal A. For the infinite case, you notice that 2^A is complete and so *there* you interpret 2^(2^A) as the set of *complete* Boolean homomorphisms from 2^A to 2. Again, you get the same result: 2^(2^A) = A, or at least it should[1]. Then you mention the role of forgetful functors and this and that (including Stone duality, which always hurt my poor addled brain). But the niggling point I don't get is the different cases for the finite and infinite sets A. When A is finite, you view 2^A as a functor from Set to BoolAlg and otherwise to CompBoolAlg, but this ad hoc change of codomain seems to strongly undercut the naturalness of your analysis, doesn't it? Without it, you lose the identity, but with it, the whole "should" seems much weaker: it *should* only if 2^(2^A) retains those features of 2^A which come from A's being finite/infinite[2]. But why those features and not other features of A? Or why *any* features of A per se? Why not just BoolAlg? Anyway, I hope I haven't missed the point entirely. Thanks for any comments. Footnotes: [1] I don't know the proper notation for "should =", but I suppose that it is O(x = y) (where O is the deontic circle). [2] Is there a term like "parity" which applies to the finite/infinite dichotomy? -- "I am a force of Nature. Time is a friend of mine, and We talk about things, here and there. And sometimes We muse a bit [...] and then We watch them go... in the meantime, Time and I, We play with some of them, at least for a little while." --- JSH and His pal, Time.
From: Chris Menzel on 30 Jul 2005 23:27
On Wed, 27 Jul 2005 23:02:26 -0700 (PDT), Alan Morgan <amorgan(a)xenon.Stanford.EDU> said: > In article <MPG.1d51c4219426ae86989fd5(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >>Daryl McCullough said: >>> Tony Orlow says... >>> >>> >despite the fact that an infinite set of whole numbers requires >>> >infinite whole numbers >>> >>> That's false, no matter how many times you say it. No finite >>> set can contain every (finite) natural. Why? Because every finite >>> set of naturals has a largest element, and there is no largest >>> finite natural. >>That's false, no matter how many times you say it. > > Which part do you disagree with? Do you disagree that every finite > set of naturals has a largest element or do you disagree that there > is no largest finite natural? Don't expect an answer. Daryl has pointed out this overt contradiction in TO's "views" in several posts, all ignored. It gets much too clearly and succinctly to the heart of the matter. |